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vodka [1.7K]
2 years ago
12

An orchard produced 526.96 liters of fresh-squeezed orange juice over the course of 6 days. To the nearest thousandth of a liter

, how much orange juice was produced each day, on average?
Mathematics
1 answer:
Alona [7]2 years ago
7 0

Answer:

Hello!!

If an orchard produced 526.96 liters of fresh-squeezed orange juice over the course of 6 days. To the nearest thousandth of a liter, there was 87.827 liters of juice produced each day on average.

Step-by-step explanation:

526.96 ÷ 6=87.8266667=87.827

Hope this helps!!

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A well known pharmaceutical manufacturer is manufacturing a newly developed vaccine, and is concerned about variability in the i
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Answer:

a

The  null hypothesis is H_o :  \  \sigma^2 =  1.9

The alternative hypothesis is  H_a :  \  \sigma^2 \ne 1.9  

b

 X^2 =  13.74

c

The decision rule is  

Reject the null hypothesis

d

The conclusion is  

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

Step-by-step explanation:

From the question we are are told that

   The variance is \sigma ^2 =  1.9

    The sample size is  n =  30  

    The  sample variance  is  s^2 = 0.9

    The level of significance is  \alpha  =  0.05

   

The  null hypothesis is H_o :  \  \sigma^2 =  1.9

The alternative hypothesis is  H_a :  \  \sigma^2 \ne 1.9  

Generally the test statistics  is mathematically represented as

          X^2 = \frac{ (n- 1 ) * s^2 }{ \sigma^2 }

=>       X^2 = \frac{ (30 - 1 ) * 0.9 }{1.9  }

=>       X^2 =  13.74

Generally  from the degree of freedom is mathematically represented as

         df =  n- 1

=>      df =   30 - 1

=>      df =   29

Generally from the chi distribution table the critical value of  \frac{\alpha}{2}  \   and  \   1 - \frac{ \alpha }{2} at a degree of freedom  of  df =   29 is  

       X^2 _{ \frac{\alpha }{2} ,  df } =  X^2 _{ \frac{0.05 }{2}  , 29 }  =  45.7

=>     X^2 _{ 1 - \frac{\alpha }{2} ,  df } =  X^2 _{1 -  \frac{0.05 }{2}  , 29 }  =  16.0 5

Gnerally from the values obtained we see that

         X^2 <  X^2 _{ 1 - \frac{\alpha }{2}  ,  df } hence

The decision rule is  

Reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

6 0
3 years ago
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